Question: Multiply the following complex numbers: $({5+2i}) \cdot ({-4-i})$
Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({5+2i}) \cdot ({-4-i}) = $ $ ({5} \cdot {-4}) + ({5} \cdot {-1}i) + ({2}i \cdot {-4}) + ({2}i \cdot {-1}i) $ Then simplify the terms: $ (-20) + (-5i) + (-8i) + (-2 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ -20 + (-5 - 8)i - 2i^2 $ After we plug in $i^2 = -1$ , the result becomes $ -20 + (-5 - 8)i - (-2) $ The result is simplified: $ (-20 + 2) + (-13i) = -18-13i $